Constant force acting on a particle

In summary, the conversation discusses a projectile motion problem and concludes that the path is parabolic. The equations of trajectory are given but the solution is not clear. The conversation then suggests to eliminate the quadratic terms in ##t## by calculating ##3x+4y## and solving for ##t##. After substituting ##t##, a parabola tilted at 45 degrees is obtained, but with a term containing ##xy## which is unusual for a parabola. It is explained that this indicates the curve is rotated at an angle from the ##x## or ##y## axis.
  • #1
Vibhor
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Homework Statement


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Homework Equations

The Attempt at a Solution



If I consider a coordinate system with the -Y axis along the direction of force and X axis along a line perpendicular to it (except the direction of velocity vector) then this problem is equivalent to the usual projectile motion problem where a particle is projected from an elevation horizontally ( i.e velocity perpendicular to gravity ) .

So I can conclude that the path in the problem is parabolic i.e option b) which is indeed the given answer .

Now , when I tried to find the equation of trajectory , I am a bit lost .

In the usual X-Y coordinate system with 'm' representing the mass of the particle , we have ,

## x = 3t + \frac{1}{2}\frac{4}{m}t^2##

## y = 4t - \frac{1}{2}\frac{3}{m}t^2##

I am finding it difficult to solve the above equations so as to get the relation between x and y .

Any sincere help is very much appreciated .

Thanks .
 

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  • #2
Calculate ##3x+4y## to eliminate the quadratic terms in ##t##. Then solve this for ##t## and substitute for ##t## in either ##x(t)## or ##y(t)##. You should get a parabola tilted at 45 degrees.
 
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  • #3
blue_leaf77 said:
Calculate ##3x+4y## to eliminate the quadratic terms in ##t##. Then solve this for ##t## and substitute for ##t## in either ##x(t)## or ##y(t)##. You should get a parabola tilted at 45 degrees.

Fantastic ! :smile:

But on expanding I get a term containing ##xy## . I have never really seen an ##xy## term in the usual parabola equations .
 
  • #4
Vibhor said:
Fantastic ! :smile:

But on expanding I get a term containing ##xy## . I have never really seen an ##xy## term in the usual parabola equations .
The coupled terms like that in conic sections such as parabola is an indication that this curve is rotated at an angle from ##x## or ##y## axis.
 
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  • #5
blue_leaf77 said:
The coupled terms like that in conic sections such as parabola is an indication that this curve is rotated at an angle from ##x## or ##y## axis.

Interesting :cool: .

Thanks a lot .
 

1. What is a constant force acting on a particle?

A constant force acting on a particle is a force that remains the same in magnitude and direction throughout the entire duration of the particle's motion. It does not change or vary in any way.

2. How does a constant force affect the motion of a particle?

A constant force will cause a particle to accelerate in the direction of the force. The acceleration will be directly proportional to the force and inversely proportional to the mass of the particle.

3. Can a constant force change the direction of a particle's motion?

Yes, a constant force can change the direction of a particle's motion if it is not acting in the same direction as the particle's initial velocity. This will cause the particle to change its velocity and therefore its direction of motion.

4. What is the formula for calculating the acceleration of a particle under a constant force?

The formula for calculating the acceleration of a particle under a constant force is a = F/m, where a is the acceleration, F is the force, and m is the mass of the particle.

5. Can a constant force cause a particle to have a constant velocity?

No, a constant force will always cause a particle to accelerate and therefore have a changing velocity. In order for a particle to have a constant velocity, there must be no net force acting on it.

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